I have the following verb defintions, which produce
two slightly different results: there are extra spaces in
the `;@boxedtoken` result, which I want to re-introduce into
a later spinoff of a result that resembles the `lr` result
when both are applied to the same argument..
lr =: 5!:5&<
boxedtoken =: ;:@lr
For example, consider the verb fetch defined as
follows in the dictionary d621.htm .
fetch=: >@({&>/)@(<"0@|....@[ , <@]) " 1 _
I hope you can see the extra spaces by looking at
the following results.
,.(;@boxedtoken ;lr) 'fetch'
+-----------------------------+
|>@({&>/)@(<"0@|....@[,<@])"1 _ |
+-----------------------------+
|>@({&>/)@(<"0@|....@[ , <@])"1 _|
+-----------------------------+
,.(;@boxedtoken ;&(' '&=)lr) 'fetch'
+---------------------------------------------------------+
|0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 |
+---------------------------------------------------------+
|0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0|
+---------------------------------------------------------+
How can I generate a binary list that can be applied
to the result of `;@boxedtoken 'fetch'` to expand it by
inserting the two missing spaces so that it matches the
result of `lr 'fetch'`?
To simplify: For example if the two results were
"abc" and "a bc" then the mask would be "1j1 1 1" because as
shown in the following, the expanded result is produced.
1j1 1 1#'abc'
a bc
The correct expansion list for my example is the
following, but I am looking for a code snippet which uses
the results of `;@boxedtoken 'fetch'` and `lr 'fetch'` to
produce the resulting list.
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1j1 1j1 1 1 1 1 1 1 1 1
Thanks,
--
(B=)
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